A combination is simply a selection, where the order of selection is not important.
Choosing strawberries and ice cream from a menu is the same combination as choosing ice cream and strawberries.
When we select r objects in no particular order from n objects, we call this is a combination.
A combination of r objects which are then arranged in order is equivalent to a
permutation. We write nCr to mean the number of combinations of r objects from n. Since there are rPr = r! ways of arranging the r objects, we have:
C P P
C r n
n r
C n
r n r
n r r
r n
r n r
n r
! !
( )!
!
!( )!
× =
× = −
= −
Suppose we wish to select three children from a group of five. We can view this task as ‘choosing three and ignoring two’ or as ‘choosing to ignore two and remaining with three’. Regardless of how we view it, choosing three from five and choosing two from five can be done in an equal number of ways, and so 5C3=5C2.
The following three points should be noted.
C C
n r n
n r–
= C ø P
n r n
r
C n
r n r
n r
!
!( )!
No. we select from!
No. selected! No. not selected!
= − =
×
n r
n r n r
!
!( )!
=
− We will use this more modern notation in Chapter 7. However, most calculators use the
nC
r notation, so we will use this in the current chapter.
FAST FORWARD There are
C n
r n r
n r
!
!( )!
= −
combinations of r objects from n distinct objects.
KEY POINT 5.5
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How many distinct three-digit numbers can be made from five cards, each with one of the digits 5, 5, 7, 8 and 9 written on it?
Answer
The 5 is a repeated digit, so we must investigate three situations separately.
No 5s selected: 3P3=6 three-digit numbers.
The digits 7, 8 and 9 are selected and arranged.
One 5 selected: 3C2×3!=18 three-digit numbers.
Two digits from 7, 8 and 9 are selected and arranged with a 5. Two 5s selected: C 3!
2! 9
3 1× =
three-digit numbers.
One digit from 7, 8 and 9 is selected and arranged with two 5s.
6+18+ =9 33 three-digit numbers can be made.
WORKED EXAMPLE 5.14
The selections in these three situations are mutually exclusive, so we add together the numbers of three-digit numbers.
TIP
EXERCISE 5F
1 Find the number of ways in which five apples can be selected from:
a eight apples b nine apples and 12 oranges.
2 From seven men and eight women, find how many ways there are to select:
a four men and five women b three men and six women c at least 13 people.
3 a How many different hands of five cards can be dealt from a standard deck of 52 playing cards?
b How many of the hands in part a consist of three of the 26 red cards and two of the 26 black cards?
4 a From the 26 letters of the English alphabet, find how many ways there are to choose:
i six different letters ii 20 different letters.
b Use your results from part a to find the condition under which xCy =xCz, where x is a positive integer.
You will learn about probability distributions for the number of objects that can be selected in Chapter 6, such as the number of women selected for this team.
FAST FORWARD A team of five is to be chosen from six women and five men. Find the number of
possible teams in which there will be more women than men.
Answer From 6 women
From 5 men
No. teams
3 2 6C3 5C 200
× 2=
or 4 1 6C4×5C1=75
or 5 0 6C5 5C 6
× 0=
200+75+ =6 281 teams with more women than men.
The table shows the possible make-up of the team when it has more women than men in it; and also the number of ways in which those teams can be chosen.
WORKED EXAMPLE 5.13
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5 In a classroom there are four lights, each operated by a switch that has an on and an off position. How many possible lighting arrangements are there in the classroom?
6 From six boys and seven girls, find how many ways there are to select a group of three children that consists of more girls than boys.
7 A bag contains six red fuses, five blue fuses and four yellow fuses. Find how many ways there are to select:
a three fuses of different colours b three fuses of the same colour c 10 fuses in exactly two colours d nine fuses in exactly two colours.
8 The diagram opposite shows the activities offered to children at a school camp.
If children must choose three activities to fill their day, how many sets of three activities are there to choose from?
9 Two taxis are hired to take a group of eight friends to the airport. One taxi can carry five passengers and the other can carry three passengers.
What information is given in this situation by the fact that 8C5=8C3=56?
10 Ten cars are to be parked in a car park that has 20 parking spaces set out in two rows of 10. Find how many different patterns of unoccupied parking spaces are possible if:
a the cars can be parked in any of the 20 spaces b the cars are parked in the same row
c the same number of cars are parked in each row d two more cars are parked in one row than in the other.
11 A boy has eight pairs of trousers, seven shirts and six jackets. In how many ways can he dress in trousers, shirt and jacket if he refuses to wear a particular pair of red trousers with a particular red shirt?
12 A girl has 11 objects to arrange on a shelf but there is room for only seven of them.
In how many ways can she arrange seven of the objects in a row along the shelf, if her clock must be included?
13 A Mathematics teacher has 10 different posters to pin up in their classroom but there is enough space for only five of them. They have three posters on algebra, two on calculus and five on trigonometry. In how many ways can they choose the five posters to pin up if:
a there are no restrictions
b they decide not to pin up either of the calculus posters
c they decide to pin up at least one poster on each of the three topics algebra, calculus and trigonometry?
14 As discussed at the beginning of this chapter in Explore 5.1 about encrypting letters, it states that there are over 27 million possibilities for the password encrypted as UJSNOL. How many possibilities are there?
15 How many distinct three-digit numbers can be made from 1, 2, 2, 3, 4 and 5, using each at most once?
Today’s Activities Morning: acting, painting or singing Afternoon: swimming, tennis, golf or cricket Evening: night-hike, star-gazing or drumming Afternoon swimming can be done at the pool or at the lake
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Two women and three men can sit on a five-seater bicycle in 5!=120 different ways. The photo shows an arrangement in which the two women are separated and the three men are also separated.
Consider, separately, the arrangements in which the women, and in which the men, are all separated from each other.
a Women separated from each other.
Women next to each other =2!
Arrange three men with the women as a single object =4!
There are 2!×4! arrangements in which the women are not separated.
So there are 5!−(2!×4!)=72
arrangements in which the women are separated from each other.
b Men separated from each other.
Men next to each other =3!
Arrange two women with the men as a single object =3!
There are 3!×3! arrangements in which the men are not separated.
So there are 5! – (3!×3!)=84 arrangements in which the men are separated from each other.
The calculations in a and b follow the same steps; however, the logic in one of them is flawed. Which of the two answers is correct? Can you explain why the other answer is not correct?
EXPLORE 5.3